\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F_{newton}\)
then effectively
\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o.c}=F_{newton}\)
that
\(\varepsilon_o\rightarrow \varepsilon_o^{*}.c\)
in which case,
\({\varepsilon_o^{*}.c}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\)
\({\varepsilon_o^{*}}=\cfrac{1}{2c^3ln(cosh(\theta_{\psi}))}\)
\(c=\cfrac{1}{\sqrt[3]{2ln(cosh(\theta_{\psi})).\varepsilon_o^{*}}}=\cfrac{1}{\sqrt{\mu_o\varepsilon_o^{*}}}=\cfrac{1}{\sqrt{\varepsilon_o2ln(cosh(\theta_{\psi}))}}\) --- (*)
\(\mu_o\) in the denominator is first cube rooted then squared then divided by \(2ln(cosh(\theta_{\psi}))\). The far right hand side, is the actual measured term, and has the factor \(2ln(cosh(\theta_{\psi}))\). The mistaken theoretical term involving \(\varepsilon_o^{*}\) does not, so \(\mu_o\) inserted into the expression is divided by \(2ln(cosh(\theta_{\psi}))\).
With the result,
\({\varepsilon_o}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\)
when \(\varepsilon_o\rightarrow \varepsilon_o^{*}.c\), we have
\({\varepsilon_o^{*}.c}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\)
\({\varepsilon_o^{*}.c}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\)
Multiplied by \(\mu_o\),
\({\mu_o\varepsilon_o^{*}.c}=\mu_o.\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}.c\)
if this were to be
\(\mu_o\varepsilon_o=\cfrac{1}{c^2}\)
\(\mu_o.\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}.c=\cfrac{1}{c^2}\)
\(\mu_o=\cfrac{2ln(cosh(\theta_{\psi}))}{c}\)
such that we move the measured results to the mistaken theoretical expression of \(\cfrac{1}{c^2}\).
such that we move the measured results to the mistaken theoretical expression of \(\cfrac{1}{c^2}\).
So when we apply information in (*) to the \(\mu_o\) above,
\(\mu_o=\cfrac{1}{2ln(cosh(\theta_{\psi}))}\left(\sqrt[3]{\cfrac{2ln(cosh(\theta_{\psi}))}{c}}\right)^2\)
\(\mu_o=\left(\cfrac{1}{c\sqrt{2ln(cosh(\theta_{\psi}))}}\right)^{2/3}\)
when \(\theta_{\psi}=3.135009\),
\(\mu_o=1.315500e-6\)
The defined value of \(\mu_o\) is \(\mu_o=1.256637e-6\)
Which is interesting. \(\mu_o\) may have originated in the mistake in \(c\) in,
\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F.c\ne F_{newton}\)
